Rotor Blades Construction

09 Jan

Author: admin | Category: Gyrocopters For Sale

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Rotor Blades Construction

physics: dynamics, momentum and newton's laws of motion?

(a)Some makes of car have, as a safety feature, regions at the front and rear which are designed to collapse on impact, but the shell of the passenger compartment is of rigid construction. Give a brief physical explanation of how this design may help to protect passengers from serious injury in
the event of a collision.

(b)Is a force of 168N exerted on the neck big enough to cause serious injury?

(c)A model helicopter of mass 5.0 kg rises with constant acceleration from rest to a height of 60 m in 10 s. Find the thrust exerted by the rotor blades during the ascent.

(a) An impact is an impulsive force; qualitatively, it is a force which is intense for a very short period of time. In the event of an impact, the regions that are designed to collapse (and thus protect the passengers in the car) are made of such materials and of such a construction that this impulsive force has its time of action extended over a longer period of time than if this region were made of rigid material. Thus the energy that is involved in the impact is transmitted into these regions, deforming them, thus slowly dissipating this energy in so doing and, ideally, at the inner compartments of the car - where the passengers are - the residue of this energy is zero.

(b) It is almost certain that a force of this magnitude, if applied as an impulse-force, would sever the entire neck of the poor unfortunate victim! However, if this same force was applied gradually, but the neck was restrained so that it could not move when this force was applied, a crushing force would then develop - with the same unfortunate result. However, if the neck (and, presumably, the entire victim also!) was free to move, without restriction, when such a force was applied gradually, very little harm - if any - would befall the man's neck. An illustration of this is an astronaut in a space ship at take-off.

(c) We invoke the standard formula for constant acceleration: s = u.t + ½f.t²,
where: u = initial speed; v = final speed ater time t & the distance travelled is s.

In this helicopter example, u = 0 (starts from rest)
Hence s = 60 = ½.f.(10)² . . . . . . so f = 1.2 m/s².

Invoking Newton's 2nd. Law of Motion, to wit:
P = m.f, where m = mass; f = acceleration and P = force,
Here, P = 5.0 Kg. times 1.2 m/s² = 6 Newtons. This is the thrust developed by the rotor blades on the surrounding air.

N.B. It is illustrative to compare this lifting force (6 Newtons) to the specified 168 Newton's in part (b) !